A finite difference approximation to the solution of the two-point boundary value problem y″ = f (x)y + g(x), x ∈[a, b] y(a) = A, y(b) = B is defined by – h^–2 (yn–1 – 2yn + yn+1) + f (xn)yn = – g(xn), 1 ≤ n ≤ N – 1, and y0 = A, yN = B, where N is an integer greater than

Question

A finite difference approximation to the solution of the two-point boundary value problem

[latex] \begin{aligned} & y^{\prime \prime}=f(x) y+g(x), x \in[a, b] \ & y(a)=A, y(b)=B \end{aligned} [/latex]

is defined by

[latex]-h^{-2}\left(y_{n-1}-2 y_{n}+y_{n+1} ight)+f\left(x_{n} ight) y_{n}=-g\left(x_{n} ight), 1 \leq n \leq N-1,[/latex]

and [latex] y_0=A, y_N=B, [/latex]

where [latex]N[/latex] is an integer greater than [latex]1, h=(b-a) / N, x_{n}=a+n h[/latex], and [latex]y_{n}[/latex] denotes the approximation to [latex]y\left(x_{n} ight)[/latex].

(i) Prove that if [latex]f(x) \geq 0, x \in[a, b][/latex] and [latex]y(x) \in C^{4}[a, b][/latex], then

[latex]\left|y\left(x_{n} ight)-y_{n} ight| \leq \frac{h^{2}}{24} M_{4}\left(x_{n}-a ight)\left(b-x_{n} ight)[/latex]

where [latex]M_{4}=\max _{x \in[a, b]}\left|y^{(4)}(x) ight|.[/latex]

(ii) Show that with [latex]N=3[/latex], the difference scheme gives an approximation to the solution of

[latex] \begin{aligned} y^{\prime \prime}-y & =1, x \in[0,1] \ y(0) & =0, y(1)=e-1, \end{aligned} [/latex]

for which [latex]\left|y\left(x_{n} ight)-y_{n} ight| \leq \frac{e}{864}, 0 \leq n \leq 3[/latex].

Accepted Answer

(i) The difference equation at [latex]x=x_{n}[/latex] is defined by

[latex]-y_{n-1}+2 y_{n}-y_{n+1}+h^{2} f_{n} y_{n}=-g_{n} h^{2}, n=1(1) N-1 .[/latex]

Incorporating the boundary conditions [latex]y_{0}=A[/latex] and [latex]y_{N}=B[/latex] into the difference equations, we write the system of equations in matrix notation as

[latex] J y +h^2 F y = D [/latex]

[latex] \begin{aligned} & ext { where } \quad J =\left[\begin{array}{rrrr} 2 & -1 & & 0 \ -1 & 2 & -1 & \ & \ldots & & \ & -1 & 2 & -1 \ 0 & & -1 & 2 \end{array} ight], \quad F =\left[\begin{array}{llll} f_1 & & & 0 \ & f_2 & & \ 0 & & \ddots & \ & & & f_{N-1} \end{array} ight] \ & y =\left[\begin{array}{lllll} y_1 & y_2 \ldots y_{N_{-1}} \end{array} ight]^T, D =\left[\begin{array}{llll} A -h^2 g_1 & -h^2 g_2 & \ldots & B-h^2 g_{N_{-1}} \end{array} ight]^T . \ & \end{aligned} [/latex]

Exact solution satisfies the equation

[latex]\mathrm{J} \mathrm{y}\left(x_{n} ight)+h^{2} \mathrm{F} \mathrm{y}\left(x_{n} ight)=\mathrm{D}-\mathrm{T}[/latex]

where [latex]\mathrm{T}=\left[\begin{array}{llll}T_{1} & T_{2} & \ldots & T_{N-1}\end{array} ight]^{T}[/latex] is the truncation error.

In order to find the error equation, we put [latex]y_{n}=y\left(x_{n} ight)+\varepsilon_{n}[/latex] in the difference equation and obtain

[latex]\mathrm{J} \boldsymbol{\varepsilon}+h^{2} \mathrm{F} {\varepsilon}=\mathrm{T}[/latex]

where [latex]{\varepsilon}=\left[\begin{array}{llll}\varepsilon_{1} & \varepsilon_{2} & \ldots & \varepsilon_{N-1}\end{array} ight]^{T} .[/latex]

The truncation error is given by

[latex] \begin{aligned} T_{n} & =y\left(x_{n+1} ight)-2 y\left(x_{n} ight)+y\left(x_{n-1} ight)-h^{2} f\left(x_{n} ight) y\left(x_{n} ight)-h^{2} g\left(x_{n} ight) \ & =\frac{h^{4}}{12} y^{(4)}(\xi), x_{n-1}<\xi

Hence, [latex]\left|T_{n} ight| \leq \frac{h^{4}}{12} M_{4}, M_{4}=\max _{x \in[a, b]}\left|y^{(4)}(x) ight|.[/latex]

Since [latex]f(x) \geq 0, x \in[a, b][/latex] we have

[latex]\mathrm{J}+h^{2} \mathrm{F}>\mathrm{J}.[/latex]

The matrices [latex]\mathrm{J}[/latex] and [latex]\mathrm{J}+h^{2} \mathrm{F}[/latex] are irreducibly diagonal dominent with non-positive off diagonal elements and positive diagonal elements. Hence, [latex]\mathrm{J}[/latex] and [latex]\mathrm{J}+h^{2} \mathrm{F}[/latex] are monotone matices.

If follows that [latex]\left(\mathrm{J}+h^{2} \mathrm{F} ight)^{-1}<\mathrm{J}^{-1}[/latex].

Hence, we get [latex]\quad \boldsymbol{\varepsilon}=\left(\boldsymbol{J}+h^{2} \mathrm{F} ight)^{-1} \mathrm{T} \leq \mathrm{J}^{-1} \mathrm{T}[/latex].

We now determine [latex]\mathrm{J}^{-\mathrm{1}}=\left(j_{i, j} ight)[/latex] explicitly. On multiplying the rows of [latex]\mathrm{J}[/latex] by the [latex]j[/latex] th column of [latex]\mathrm{J}^{-1}[/latex], we have the following difference equations.

[latex] \begin{aligned} & (i)-2 j_{1, j}-j_{2, j}=0, \ & ( ext { ii })-j_{i-1, j}+2 j_{i, j}-j_{i+1, j}=0, \quad 2 \leq i \leq j-1, \ & (i i i)-j_{j-1, j}+2 j_{j, j}-j_{j+1, j}=1, \ & (i v)-j_{i-1, j}+2 j_{i, j}-j_{i+1, j}=0, \quad j+1 \leq i \leq N-2, \ & (v)-j_{N-2, j}+2 j_{N-1, j}=0. \end{aligned} [/latex]

On solving the difference equations, we get

[latex] j_{i, j}=\left|\begin{array}{ll} \frac{i(N-j)}{N}, & i \leq j, \ \frac{j(N-i)}{N}, & i \geq j, \end{array} ight. [/latex]

Note that the matrix [latex]\mathrm{J}^{-1}[/latex] is symmetric. The row sum of the [latex]n[/latex]th row of [latex]\mathrm{J}^{-1}[/latex] is

[latex]\sum_{j=1}^{N-1} j_{n, j}=\frac{n(N-n)}{2}=\frac{\left(x_{n}-a ight)\left(b-x_{n} ight)}{2 h^{2}}.[/latex]

Thus, we have

[latex] \left|\varepsilon_n ight| \leq \frac{h^4}{12} M_4 \frac{\left(x_n-a ight)\left(b-x_n ight)}{2 h^2} [/latex]

or [latex] \left|y\left(x_n ight)-y_n ight| \leq \frac{h^2}{24} M_4\left(x_n-a ight)\left(b-x_n ight). [/latex]

(ii) We are given that

[latex] \begin{aligned} & N=3, f(x)=1, g(x)=1, A=0 ,\ & B=e-1, a=0, b=1, h=1 / 3. \end{aligned} [/latex]

We have

[latex]y^{(4)}(x)=y^{\prime \prime}(x)=y(x)+1.[/latex]

Therefore, [latex] M_4=\max _{x \in[0,1]}\left|y^{(4)}(x) ight|=\max _{x \in[0,1]}|y(x)+1|=e-1+1=e . [/latex]

Maximum of [latex]\left(x_{n}-a ight)\left(b-x_{n} ight)[/latex] occurs for [latex]x_{n}=(a+b) / 2[/latex] and its maximum magnitude is [latex](b-a)^{2} / 4=1 / 4[/latex]. Hence

[latex]\left|y\left(x_{n} ight)-y_{n} ight| \leq \frac{e}{864}, 0 \leq n \leq 3.[/latex]

Question 5.70: A finite difference approximation to the solution of the two......

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